Carl Love

## 26897 Reputation

11 years, 310 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

## That's 12. What about 16?...

It's not so simple. You used 12 instead of 16. Try it with 16.

## That's 12. What about 16?...

It's not so simple. You used 12 instead of 16. Try it with 16.

## It's a sledgehammer...

@Markiyan Hirnyk The Maple approach is a sledgehammer that I don't like because it involves checking 2^15 cases to do something that should be (at least in principle) doable with secondary-school combinatorics. But I had to find out where my analysis went wrong.

## It's a sledgehammer...

@Markiyan Hirnyk The Maple approach is a sledgehammer that I don't like because it involves checking 2^15 cases to do something that should be (at least in principle) doable with secondary-school combinatorics. But I had to find out where my analysis went wrong.

## Why evalhf instead of evalf?...

I could show you how to do it with evalhf, but it is a little complicated. So, first I ask Did you mean evalf? I don't see any benefit to using evalhf in this case. So let me know if you really want evalhf, and why.

## I get it...

@Joe Riel Finally, I get it! Thank you for your patient explanation. Put more succintly (I think), in all cases I was missing the choice of whether the element(s) of (A xor B) were or were not in C. In the |AB|=4 case, there is one such element, for a factor of 2; in the |AB|=3 case, there are two such elements, for a factor of 2^2.

I editted my solution at the top of this subthread to reflect these corrections.

## I get it...

@Joe Riel Finally, I get it! Thank you for your patient explanation. Put more succintly (I think), in all cases I was missing the choice of whether the element(s) of (A xor B) were or were not in C. In the |AB|=4 case, there is one such element, for a factor of 2; in the |AB|=3 case, there are two such elements, for a factor of 2^2.

I editted my solution at the top of this subthread to reflect these corrections.

## PolynomialIdeals...

My experience with this, admittedly very limited, is that the ?PolynomialIdeals package gives you access the Groebner package with a more ideal-oriented syntax.

## You're right, I was using the wrong vers...

@Markiyan Hirnyk You're right, I was using the wrong version of Maple. I just sent you an email with more info.

## You're right, I was using the wrong vers...

@Markiyan Hirnyk You're right, I was using the wrong version of Maple. I just sent you an email with more info.

## A simulation in Maple...

@Markiyan Hirnyk By doing a simulation in Maple, I know that my answer is wrong and that Joe's edit of his original answer is correct. But I either don't understand or don't agree with Joe's analysis of what was wrong with my answer. My simulation showed that my count of the |AB|=4 case was short by a factor of 2, and my counts of both subcases of the |AB|=3 case were short by a factor of 4. Here's the simulation code:

S:= combinat[powerset](5):
n:= 0: ab3:= 0: ab4:= 0: ab5:= 0: b4:= 0:
for A in S do  for B in S do  for C in S do
if nops(A union B) = 5 and nops(A intersect B intersect C) = 3 then
n:= n+1;
abn:= nops(A intersect B);
if abn=3 then
ab3:= ab3+1;
if nops(B)=4 then b4:= b4+1 end if
elif abn=4 then ab4:= ab4+1
end if
end if
end do  end do  end do:
n, ab3, ab4, b4;

So, the final answer is 250*binomial(30,5)*2^25, which is the same as what Joe got, although he expressed it differently.

## A simulation in Maple...

@Markiyan Hirnyk By doing a simulation in Maple, I know that my answer is wrong and that Joe's edit of his original answer is correct. But I either don't understand or don't agree with Joe's analysis of what was wrong with my answer. My simulation showed that my count of the |AB|=4 case was short by a factor of 2, and my counts of both subcases of the |AB|=3 case were short by a factor of 4. Here's the simulation code:

S:= combinat[powerset](5):
n:= 0: ab3:= 0: ab4:= 0: ab5:= 0: b4:= 0:
for A in S do  for B in S do  for C in S do
if nops(A union B) = 5 and nops(A intersect B intersect C) = 3 then
n:= n+1;
abn:= nops(A intersect B);
if abn=3 then
ab3:= ab3+1;
if nops(B)=4 then b4:= b4+1 end if
elif abn=4 then ab4:= ab4+1
end if
end if
end do  end do  end do:
n, ab3, ab4, b4;

So, the final answer is 250*binomial(30,5)*2^25, which is the same as what Joe got, although he expressed it differently.

## Code Edit Region...

The best way to enter multi-line code into a Maple worksheet is to use a Code Edit Region. Place your cursor where you want the window, then select Code Edit Region from the Insert menu. It gives you a Notepad-like editor with tabs and hanging indents. Also, you can use ordinary Enter (or Return) instead of Shift-Enter (Shift-Return).

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